Three concentric spherical shells have radii a, b and c (a < b < c) and have surface charge densities \sigma , -\sigma and \sigma respectively. If V_A, V_B and V_C denote the potential of the three shells, if c = a + b, we have

1. Charge Calculation: Let $q_A, q_B, q_C$ be charges on shells of radii $a, b, c$ with densities $\sigma, -\sigma, \sigma$ respectively. $q_A = 4\pi a^2 \sigma$, $q_B = -4\pi b^2 \sigma$, $q_C = 4\pi c^2 \sigma$.

2. Potential Formulas: The potential $V$ at the surface of a shell of radius $R$ is due to its own charge and the charges of other shells. Potential inside a shell is constant and equals the surface potential. Potential outside varies as $1/r$. $V = \frac{1}{4\pi\varepsilon_0} \sum \frac{q}{r}$

3. Potential at Shell A ($V_A$): $V_A = \frac{1}{4\pi\varepsilon_0} [\frac{q_A}{a} + \frac{q_B}{b} + \frac{q_C}{c}]$ $V_A = \frac{1}{4\pi\varepsilon_0} [\frac{4\pi a^2 \sigma}{a} - \frac{4\pi b^2 \sigma}{b} + \frac{4\pi c^2 \sigma}{c}] = \frac{\sigma}{\varepsilon_0} (a - b + c)$

4. Potential at Shell C ($V_C$): $V_C = \frac{1}{4\pi\varepsilon_0} [\frac{q_A}{c} + \frac{q_B}{c} + \frac{q_C}{c}]$ $V_C = \frac{1}{4\pi\varepsilon_0} \frac{\sigma}{c} (4\pi a^2 - 4\pi b^2 + 4\pi c^2) = \frac{\sigma}{\varepsilon_0 c} (a^2 - b^2 + c^2)$

5. Apply Condition c = a + b: Substitute $c = a + b$ into $V_A$: $V_A = \frac{\sigma}{\varepsilon_0} (a - b + a + b) = \frac{\sigma}{\varepsilon_0} (2a)$

Substitute $c = a + b$ into $V_C$: $V_C = \frac{\sigma}{\varepsilon_0 (a+b)} (a^2 - b^2 + (a+b)^2)$ $V_C = \frac{\sigma}{\varepsilon_0 (a+b)} ((a-b)(a+b) + (a+b)^2)$ $V_C = \frac{\sigma}{\varepsilon_0 (a+b)} (a+b) [ (a-b) + (a+b) ]$ $V_C = \frac{\sigma}{\varepsilon_0} (2a)$

6. Conclusion: $V_A = V_C$. We calculate $V_B$ to check equality. $V_B = \frac{\sigma}{\varepsilon_0} (\frac{a^2}{b} - b + c) = \frac{\sigma}{\varepsilon_0} (\frac{a^2}{b} - b + a + b) = \frac{\sigma}{\varepsilon_0} (\frac{a^2}{b} + a) \neq 2a$ (since $a \neq b$). Thus, $V_C = V_A \neq V_B$.